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AP Physics 1 – Uniform Circular Motion: Centripetal Force, Banked Curves & Vertical Circles

Golden Rule: Centripetal force is not a new force. It is just a name for the net force pointing toward the center (often Tension, Friction, Normal Force, or Gravity).

1. Uniform Circular Motion: Velocity vs. Acceleration

In uniform circular motion, the speed is constant, but the velocity keeps changing because its direction continuously changes toward the tangent of the circle. This changing direction is what creates a non‑zero acceleration even when the speed stays the same.

Top-down diagram of a ball on a string showing velocity tangent to the circle and centripetal force pointing toward the center. — Velocity is **tangent** to the circle. Acceleration (and net force) points **radially inward** toward the center.

Centripetal Acceleration $a_c$

Always points toward the center of the circle.

$a_c = \dfrac{v^2}{r}$

Centripetal Force $F_c$

The net inward force required to produce $a_c$ .

$\Sigma F_{\text{toward center}} = \dfrac{mv^2}{r}$

1.1 For the Nerds: Deriving $a_c = \dfrac{v^2}{r}$

Here is a quick geometric proof using similar triangles.

Geometric derivation of centripetal acceleration using similar triangles of position and velocity vectors. — The triangle for arc length $\Delta s$ and radius $r$ is similar to the triangle for $\Delta v$ and $v$ .

The geometry gives the ratio

$\dfrac{\Delta v}{v} = \dfrac{\Delta s}{r}$

Divide both sides by $\Delta t$ :

$\dfrac{\Delta v}{\Delta t} = \dfrac{v}{r} \cdot \dfrac{\Delta s}{\Delta t}$

Since $\Delta v / \Delta t = a$ and $\Delta s / \Delta t = v$ , we obtain

$a_c = \dfrac{v \cdot v}{r} = \dfrac{v^2}{r}$

1.2 Period, Frequency & Speed on a Circle

In many AP Physics 1 circular‑motion problems, it is easier to think in terms of how long one revolution takes instead of just meters per second.

Period $T$

Time for one full revolution (seconds).

$T = \dfrac{1}{f}$

Frequency $f$

Revolutions per second (Hz).

$f = \dfrac{1}{T}$

Speed in a circle:

If one revolution travels a distance $2\pi r$ in time $T$ , then

$v = \dfrac{2\pi r}{T}$

2. Flat Curve: The “Car on a Curve” Problem

When a car turns a corner on level ground, the steering wheel only changes direction; the actual inward force is provided by static friction between the tires and the road. If friction disappears (ice), the car slides straight because no force points toward the center.

Free body diagram of a car turning left, showing friction pointing left as the centripetal force. — On a dry road, static friction provides the inward (centripetal) force. On ice, friction is nearly zero and the car slides straight.

How to Solve Flat‑Curve Problems

Set the friction force equal to the required centripetal force:

$F_f = F_c$

$\mu_s F_N = \dfrac{mv^2}{r}$

(From Unit 2: on flat ground, $F_N = mg$ , so $\mu_s mg = mv^2 / r \Rightarrow v = \sqrt{\mu_s r g}.$

2.5 Banked Turns (No Friction)

A classic AP Physics 1 question: “A car drives on a banked track. At what speed can it travel without relying on friction?” This has the same math as a conical pendulum.

The key is to resolve the Normal Force into vertical and horizontal components.

Free body diagram of a car on a banked curve showing the horizontal component of Normal Force acting as centripetal force. — The horizontal component $F_N \sin\theta$ supplies the centripetal force toward the center of the curve.

“No‑Friction” Banked Turn Derivation

Vertical (y): $F_N \cos\theta = mg$ (balances gravity)
Horizontal (x): $F_N \sin\theta = \dfrac{mv^2}{r}$ (provides centripetal force)

Divide the x‑equation by the y‑equation; mass and $F_N$ cancel:

$\tan\theta = \dfrac{v^2}{rg} \Rightarrow v = \sqrt{rg \tan\theta}$

2.6 Conical Pendulum

A small mass whirling in a horizontal circle on a string (the “flying pig”) looks scary but is mathematically identical to the banked‑curve problem.

Free body diagram of a conical pendulum showing tension broken into vertical and horizontal components. — Replace Normal Force with Tension $T$ ; the components work out the same way.

Break Tension into components:

Vertical: $T\cos\theta = mg$ (balances gravity)
Horizontal: $T\sin\theta = \dfrac{mv^2}{r}$ (provides $F_c$ )

Dividing gives the same result: $\tan\theta = \dfrac{v^2}{rg}$ .

2.7 Banked Turn with Friction

Real roads have friction, which creates a range of safe speeds rather than just one speed. The direction of friction depends on whether the car tends to slide up or down the slope.

Free body diagram of a car on a banked curve with friction pointing down the slope for maximum speed. — At maximum speed, friction points down the slope to help keep the car in the turn.

Case A: Maximum Speed

Car tends to slide OUT; friction points down the slope.

$F_c = F_N\sin\theta + f\cos\theta$

Case B: Minimum Speed

Car tends to slide IN; friction points up the slope.

$F_c = F_N\sin\theta - f\cos\theta$

3. Vertical Circles (Loops & Roller Coasters)

Vertical‑circle problems are harder because gravity can help or oppose the inward direction depending on where the object is on the path. Remember: centripetal force is always the sum of forces toward the center.

Rule: Write $\Sigma F_{\text{toward center}} = ma_c$ and pay attention to signs at the top vs bottom of the loop.

Top of the Loop

Free body diagram of a roller coaster at the top of a loop showing gravity and normal force both pointing down.

Both Gravity and Normal Force point inward (toward the center).

$F_N + F_g = \dfrac{mv^2}{r}$

Bottom of the Loop

Free body diagram of a roller coaster at the bottom of a loop showing normal force pointing up and gravity pointing down.

Normal Force points inward (up), Gravity points outward (down).

$F_N - F_g = \dfrac{mv^2}{r}$

Critical Speeds for a Loop‑the‑Loop

1. Minimum speed at the top:
Just enough to keep contact ( $T = 0$ ): $v_{\text{top}} = \sqrt{gr}$
2. Minimum speed at the bottom:
Using energy to get from bottom to top gives $v_{\text{bottom}} = \sqrt{5gr}$
(Derivation connects with conservation of energy in Unit 4.)

4. AP Physics 1 Circular Motion Practice Problems

Concept Check (MCQ Style)

Q1: A ball is swung in a horizontal circle of radius $r$ with speed $v$ . If the speed is doubled to $2v$ and radius stays the same, how does the required centripetal force change?

Click to see answer

Answer: It becomes four times larger.
Since $F_c \propto v^2$ , doubling $v$ gives $(2v)^2 = 4v^2$ , so the force is $4F_c$ .
Tip: This uses proportional reasoning from the kinematics equation sheet.